Alternating Harmonic Series

By Maclaurin series , we have:

$\begin{aligned} \ln(1+x) & = x - \frac {x^2}2 + \frac {x^3}3 - \frac {x^4}4 + \cdots & \small \color{#3D99F6} \text{Putting }x = 1 \\ \implies \boxed{\ln 2} & = 1 - \frac 12 + \frac 13 - \frac 14 + \cdots \end{aligned}$

if we calculate the Taylor series for ln(x+1), we find it is:

0 + x - (x^2)/2 + (x^3)/3 - (x^4)/4 + . . .

And substituting in 1 for x we get our original sum:

= 1 - 1/2 + . . .

And now, looking at the left hand side of the equation we see

ln(1 + (1)) = . . .

So in full the equation reads:

ln(2) = 1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + . . .

giving us our answer: ln(2)!